> [!NOTE] Lemma > Define $f:\mathbb{N}^{+}\to \mathbb{R}$ by the sum $f(n)=\sum_{i=1}^{n} x^{i} = \frac{1-x^{n+1}}{1-x}.$If $x>1,$ then $f(n)$ is [[Big Theta Relation on Real Sequence|big theta]] of $x^{n},$ i.e $f(n)=\Theta(x^{n}).$ > If $x=1,$ then $f(n)=\Theta(n).$ > If $0<x<1,$ then $f(n)=\Theta(1).$ **Proof**: ...